Q:

The U.S. Census Bureau announced that the median sales price in 2014 for new homes was $282,800 and the mean sales price was $345,800. The standard deviation was $90,000. a. If you selected a random sample of 100 homes sold, what is the probability that the sample mean would be less than $370,000? b. If you selected a random sample of 100 homes sold, what is the probability that the sample mean would be between $350,000 and $360,000?

Accepted Solution

A:
Answer:a) 0.9964b) 0.3040Step-by-step explanation:Given data:standard deviation = Β $90,000Mean sales price =$345,800sample mean = Β $370,000Total number of sample = 100calculate z score for [/tex](\bar x = 370000)[/tex][tex]z = \frac{\bar x - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex][tex]z = \frac{370000 - 345800}{\frac{90000}{\sqrt{100}}}[/tex]z = 2.689P(x<370000) = P(Z<2.689)FROM STANDARD NORMAL DISTRIBUTION TABLE FOR Z P(Z<2.689) = 0.9964B)calculate z score for (\bar x = 350000)[tex]z = \frac{\bar x - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex][tex]z = \frac{350000 - 345800}{\frac{90000}{\sqrt{100}}}[/tex]z = 2.133[tex]P(350000 \leq \bar x \leq 365000) = P(0.467 \leq Z\leq 2.133) = P(Z\leq 2.133) - P(Z\leq 0.467)[/tex]FROM NORMAL DISTRIBUTION TABLE Z VALUE FOR [tex]P(Z\leq 2.133) = 0.9836[/tex][tex]P(Z\leq 0.467) = 0.6796[/tex]SO, Β = 0.9836 - 0.6796 = 0.3040